Modern Algebra I Section 1 Assignment 6 · JOHN PERRY
Exercise 1. (pg. 64 Warm Up a) Why is a linear polynomial in Q[x] always irreducible? Solution: If the linear polynomial p was not irreducible, then there would be two polynomials a, b ∈ [x] such that ab = p and dega,deg b 1. But then 1 = deg p = dega +deg b 2 by Theorem Q ≥ ≥ 4.1, which is a contradiction. ◊ 2 Exercise 2. (pg. 64 Warm Up b) Why is a polynomial of the form x + a Q[x], where a > 0, always irreducible? ∈ Solution: If the polynomial were not irreducible, then there would be two polynomials a, b Q[x] 2 ∈ such that x + a = ab and dega,deg b 1. By Theorem 4.1, dega = deg b = 1. So a and b are ≥ linear polynomials, say px q and r x s say, for p, q, r, s Q. By the Root Theorem (4.3), 2 − 2 − ∈ 2 that means (q/p) + a = 0 and (s/r ) + a = 0. But it is easy to see that b + a a > 0 and 2 ≥ c + a a > 0, a contradiction. ≥ ◊ 4 2 Exercise 3. (pg. 65 Warm Up c) Determine a factorization of x 5x +4 into irreducibles in [x]. − Q Solution: Using the Rational Root Theorem (5.6), we know that all rational roots of the polynomial are of the form 4 2 1 = 4 = 2 = 1. ±1 ± ± 1 ± ± 1 ± If we apply the Root Theorem (4.3) and substitute these into the polynomial, we find that 1, -1, 2, and -2 are roots of the polynomial. Thus it factors as (x + 1)(x 1)(x + 2)(x 2). − − ◊ Exercise 4. (pg. 65 Warm Up e) We know that 7 is an irreducible integer, but is 7 an irreducible polynomial? Solution: The definition of an irreducible polynomial p includes the requirement that deg p > 0. Since deg7 = 0, 7 is not an irreducible polynomial. ◊ 3 2 Exercise 5. (pg. 65 Warm Up g) Factor 2x + 7x 2x 1 completely into irreducibles in Q[x], using Gauss’ Lemma and the Root Theorem. Adjust your− factorization− (if necessary) so that all factors belong to Z[x]. Solution: We will also use the Rational Root Theorem (5.6)—I have no idea why the book didn’t suggest that. 1 Modern Algebra I Section 1 Assignment 6 John Perry 2 · · By the Rational Root Theorem, we know that all rational roots of the polynomial are of the form 1 1 . ±2 ± 1 If we apply the Root Theorem (4.3) and substitute these into the polynomial, we find that 1/2 is the only rational root. We can use long division or synthetic division to show that 1 3 2 2 2x + 7x 2x 1 = x 2x + 8x + 2 . − − − 2 We can adjust the factorization using associates: factor the scalar 2 from the quadratic polynomial and distribute it into the linear polynomial. Thus 3 2 2 2x + 7x 2x 1 = (2x 1) x + 4x + 1 . − − − It is easy to use the Rational Root Theorem (5.6) and the Root Theorem (4.3) to show that this polynomial has no roots in Z. By Gauss’ Lemma (5.5) the quadratic polynomial is thus irreducible over Q. ◊ Exercise 6. (pg. 65 Exercise 1) Prove Theorem 5.1: A polynomial in Q[x] of degree greater than zero is either irreducible or the product of irreducibles. Solution: Let f Q[x] be arbitrary, but fixed. Assume deg f > 0. We proceed by induction on n = deg f . ∈ For the inductive base, let n = 1. Then f is irreducible by Exercise 1 (Warm Up a). We have shown that the assertion is true for n = 1. Assume that n > 1. For the inductive hypothesis, assume that every polynomial of degree i is either irreducible or the product of irreducibles, for all i : 1 i < n. If f is irreducible, then we are done. ≤ Otherwise, we can write f = pq for some p, q [x] where deg p,deg q > 0. Recall from ∈ Q Theorem 4.1 that deg f = deg p + deg q. Thus deg p = deg f deg q < deg f . Similarly deg q < deg f . By the inductive hypothesis, p is either irreducible or− the product of irreducibles. So we can write p = p1 p2 pr for some r and for some irreducible pi [x], for all i : 1 i r . ··· ∈ N ∈ Q ≤ ≤ By similar reasoning, we can write q = q1q2 qs for some s and for some irreducible ··· ∈ N qj Q[x], for all j : 1 j s. Hence f = p1 p2 pr q1q2 qs where r, s N and pi , qj are∈ irreducible for all i :≤ 1 ≤i r and for all j : 1 ···j s. So f is··· the product of irreducibles.∈ We have shown that f is≤ either≤ irreducible or the≤ product≤ of irreducibles. Since n is arbitrary, this is true regardless of the degree of f . Since f is arbitrary, it is true for all f [x]. ∈ Q ◊ Exercise 7. (pg. 65 Exercise 4) Use Gauss’s Lemma to determine which of the following are irre- ducible in Q[x]: 3 3 2 3 2 4x + x 2, 3x 6x + x 2, x + x + x 1. − − − − Solution: It is fairly easy to use factoring by grouping and see that the second polynomial is not irre- ducible in Z[x], since 3 2 2 3x 6x + x 2 = 3x (x 2) + (x 2) − − − 2 − = (x 2) 3x 2 . − − Modern Algebra I Section 1 Assignment 6 John Perry 3 · · By Gauss’s Lemma (5.5), the polynomial is also not irreducible in Q[x]. If the other two polynomials factor, then one of the factors is linear. (See the discussion in the text on page 62.) For the third polynomial, the Root Theorem (4.3) and the Rational Root Theorem (5.6) tell us that any linear factor has the form x 1. Since neither 1 nor -1 are roots of the polynomial, it must be irreducible. For the first polynomial,± the same theorems tell us that any linear factor has one of the forms x 1, x 2,2x 1,4x 1. Exhaustive inspection (via ± ± ± ± synthetic division) shows that none of these factors the polynomial, so it is also irreducible. ◊ 4 2 Exercise 8. (pg. 65 Exercise 5) Show that x + 2x + 4 is irreducible in Q[x]. Solution: For convenience, we denote the polynomial as p. For any rational number a/b, then 4 2 (a/b) + 2(a/b) + 4 0 + 2 0 + 4 = 4 > 0, ≥ · so p has no rational roots. By the Root Theorem (4.3), it has no linear factors in Q[x]. This would exclude cubic factors as well, since if p = q r for some q, r [x], and if deg q = 3, then ∈ Q by Theorem 4.1 deg r = deg p deg q = 4 3 = 1. Since p has no linear factors, r cannot have degree 1, so q cannot have degree− 3. − It remains to show that the polynomial has no quadratic factors. Assume to the contrary that 4 2 2 2 p = x + 2x + 4 has quadratic factors p = q r . So q = ax + b x + c and r = d x + e x + f where a, b, c, d, e, f Q. By Gauss’ Lemma (5.5), we can assume q, r Z[x], so that a, b, c, d, e, f Z. If we multiply∈q, r , we can collect like terms to obtain ∈ ∈ 4 3 2 p = q r = ad x + (ae + b d) x + (a f + b e + cd) x + (b f + ce) x + c f . Two polynomials are equal iff their coefficients are equal, so
1 = ad 0 = ae + b d 2 = a f + b e + cd 0 = b f + ce 4 = c f . Since a, d are integers and ad = 1, it must be that a = d = 1. The system now becomes 0 = e + b 2 = f + b e + c 0 = b f + ce 4 = c f . Observe that b = e, so we have − 2 (1) 2 = f b + c − (2) 0 = b f b c − 4 = c f .(3) From equation (2), we know that b = 0 or f = c. We consider two cases. Modern Algebra I Section 1 Assignment 6 John Perry 4 · · Case 1: If f = c, equation (3) tells us that c = 2. Substituting this into equation (1) we see 2 2 ± that b = 2 or b = 4, neither of which has an integer solution. Since b must be an integer, − f = c. 6 Case 2: If b = 0, equation (1) tells us that f + c = 2, or f = 2 c. Substituting into equation (3), we have −
4 = c (2 c) − 2 4 = 2c c 2 − c 2c + 4 = 0. − The quadratic formula shows that this has no integer solution for c. Since c must be an integer, b = 0. 6 Neither case gives a solution for the coefficients. Hence f cannot factor as the product of two quadratic polynomials. Thus f is irreducible in Z[x]. By Gauss’ Lemma (5.5), f is irreducible in Q[x]. ◊ Exercise 9. (pg. 65 Exercise 7) Use the Rational Root Theorem 5.6 to factor
3 2 2x 17x 10x + 9. − − Solution: By the rational root theorem, the only roots of the polynomial possible are 9 9 3 3 1 1 = 9, , = 3, , = 1, . ±1 ± ±2 ±1 ± ±2 ±1 ± ±2 Using substitution (or division), we find that the roots are 9, -1, and 1/2. The polynomial factors as 1 2 x (x + 1)(x 9). − 2 − (You may obtain a different expression; I’m trying to emphasize that 1/2, 1, and 9 are all roots. − ◊ Exercise 10. (pg. 66 Exercise 8) Use the Rational Root Theorem 5.6 to argue that
3 x + x + 7
is irreducible over Q[x]. Use elementary calculus to argue that this polynomial does have exactly one real root. Solution: If the polynomial factors as pq, then by Theorem 4.1, 3 = deg p +deg q. Since this is a sum of integers, either deg p = 1 or deg q = 1. By the Rational Root Theorem (5.6), all roots would be of the form 1 7 = 1, = 7. ±1 ± ±1 ± However, none of those are roots. By the Root Theorem (4.3), the polynomial has no linear factors. Since the polynomial has no linear factors, deg p = 1 and deg q = 1. So the polynomial 6 6 cannot factor. ◊ Modern Algebra I Section 1 Assignment 6 John Perry 5 · · 2 Exercise 11. (pg. 66 Exercise 13) (a) Prove that the equation a = 2 has no rational solution; that is, prove that p2 is irrational. (This part is a repeat of Exercise 2.14.) n (b) Generalize part a, by proving that a = 2 has no rational solutions, for all positive integers n 2. ≥ Solution: 2 2 (a) If a = 2 has a rational solution, then a 2 has a rational root. By the Rational Root Theorem (5.6), any such root has the form − 1 2 = 1, = 2. ±1 ± ±1 ± 2 2 Since neither of those is a root of a 2, it follows that a = 2 has no rational solution. n − n (b) If a = 2 has a rational solution, then a 2 = 0 has a rational solution, so the polynomial an 2 has a rational root. By the Rational Root− Theorem (5.6), any such root is of the form − 1 2 = 1, = 2. ±1 ± ±1 ± Is either a root of an 2? Substitute and see: − n 1 2 = 1 2 = 1 = 0 n − 2− − 6 2 2 2 2 = 2 = 0 n − ≥ − 6 (if n is even) ( 1) 2 = 1 2 = 1 = 0 − n − 2− − 6 ( 2) 2 2 2 = 2 = 0 − n − ≥ − 6 (if n is odd) ( 1) 2 = 1 2 = 3 = 0 − n − − − − 6 ( 2) 2 8 2 = 10 = 0. − − ≤ − − − 6 n n Since none of the possibilities is a root of a 2, it follows that a = 2 has no rational solution. − ◊